Question

A buffer solution is made from partially neutralizing hypochlorous acid with sodium hydroxide; the K_a of HClO is 3.5 x 10^-8. use the Henderson-Hasselbalch equation.

A). if 50 mL of 1.0 M HClO is used to make the buffer, what volume of .50 M NaOH is needed for the final pH of the buffer to be equal to the pK_a of the acid?

B). Assuming no density change upon mixing, what are the concentrations of the weak acid (HClO) and its conjugate base (ClO^-) of this buffer.

C) if 20 mL of the buffer solution made in (a) is diluted to a new volume of 200 mL, what will be the new pH? explain why.

D) Comparing the two buffer solutions made in (a) and (c), which has higher buffing capacity if equal volumes of each are used? explain why.

E). Calculate the pH of a buffer made using 30 mL of the acid and 10 mL of the base.

Answer 1

A) HClO + NaOH ---------> NaClO + H2O

at half neutralization pH = pKa

Moles of HClO = (1mol/1000ml)×50ml = 0.05 mole

Moles of NaOH required for half neutralization = 0.025

Molarity of NaOH = 0.50M

Volume of NaOH required = (1000ml/0.50mol)×0.025= 50ml

B) Moles of HClO = 0.025mol

Moles of ClO- produced = 0.025mol

Total volume = 50ml + 50ml = 100ml

[ HClO ] = (0.025mol/100ml)×1000ml = 0.25M

[ ClO- ] = ( 0.025mol/100ml ) × 1000ml = 0.25M

C) 20ml is dilted to 200ml , so 10 time dilution

[ HClO ] = 0.25M/10 = 0.025M

[ ClO- ] = 0.25M/10 = 0.025M

Henderson - Hasselbalch equation is

pH = pKa + log( [ A-] / [ HA ] )

= 7.46 + log( 0.025/0.025 )

= 7.46 + 0

= 7.46

D) Buffer capacity , = ∆n/∆pH

Buffer capacity is defined as no of moles of acid or base required to increase the 1 pH unit

solution( a) having more moles of acid and conjucate base comparing to solution (b) , so solution (a) require more acid or base to chang the pH 1 unit

Thus , solution (a) having more buffer capacity

E)